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Mechanics connected particles

  1. Apr 5, 2013 #1
    I have been stuck on this for a while cant really see how it works, I have tried drawing it several different ways of doing it. My problem is I can kind of see how it woks in my head but, seem not to be able to put pen do paper.

    Here Is the question: The van and car eventually reach a steady speed of 18 m s-1 with the rope taut when a child runs out into the road, 30 m in front of the van. The van driver brakes sharply and decelerates uniformly to rest in a distance of 27 m.
    It takes the driver of the car 1 second to react to the van starting to brake. He then brakes and the car decelerates uniformly at f m s-2, coming to rest before colliding with the van.

    (c) Find the set of possible values of f.

    mark scheme: car has (27 + 9) m in which to stop and travels 18 m in first second
    must stop from 18 m s-1 in 18 m.

    I can work it out from here no problem simple suvat. I understand the particles will collided if the car aka B travels 36m that I understand, but I cant see how it only needs to travel 18m for a collision to happen. See to me it looks like the mark scheme has ignored the rope between and use it a two separate particles.

    could someone maybe explain it a different way. I would appreciate if someone could take the time to draw a quick diagram, so I have some sort of intuition. I am really lost on how to approach these sort of questions.

    p.s like the new layout of physics forum.

    Big thanks in advance.
     
  2. jcsd
  3. Apr 5, 2013 #2

    BruceW

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    I think you forgot to say that the rope is 9m long. Assuming that is true, then surely it makes sense that the rope has been ignored from then on? The van is decelerating, so the rope is going to be slack. So why would it make any difference from then on?
     
  4. Apr 5, 2013 #3
    Okay I understand that, but where is the 18 meters coming from. I see it like this, if both particles break at same time the both cover a distance of 27m so for a collision b has to move extra 9 m to collided with particle. So after the first second the the particles are 6 m apart then if they travel for the next 2 sec together then yes it would 18 meter for a collision to happen. Is this the correct way of seeing it.

    Oh yes, the rope is 9 m long.
     
  5. Apr 5, 2013 #4

    BruceW

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    I agree that after the first second, they are 6 m apart. But where do you get 'next 2 sec' from? The way they explain it in the mark scheme is that they calculate the place where the van stops, then say the car must stop before that point. And that the car has a 1 second delay, in which he moves 18 m, so then he has only 18 m before he reaches the point where the van will stop.
     
  6. Apr 5, 2013 #5
    Okay now I think I am fully with it. What I tried to do is Is break it down section by section. I calculate that the deceleration on the van was -6ms^2 so I tried to break it up into three parts ie three seconds. But for some reason I kept think the rope was a rod or something I dont really, why I didn't engage to this. So am I right in saying this; sorry if it what you have said already, it's just clime for me. The rope is not important because obviously it goes slack so if I take the two particle as and lookin at the without the rope connecting them, I would indedde see that after one second particle has to slow down to a complete stop over a distance of 18m to prevent a collision!
     
  7. Apr 5, 2013 #6

    BruceW

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    yep. sounds good man
     
  8. Apr 6, 2013 #7
    Thanks for the help, much appreciated. Sorry about the spelling errors; it was late when I it clicked in.
     
  9. Apr 6, 2013 #8

    BruceW

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    ah, that's alright. proper spelling is for work emails and cv's haha
     
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